CODEWITHSUNDEEP
linuxbashunixawk
krishna chaitanya jinka
krishna chaitanya jinka

Reputation: 89

if condition in awk command is not working as expected in BEGIN block. correct me if am wrong

I have a command like below

kubectl get pods |
grep -v 1/1 |
grep -v 2/2 |
awk '{print $1:$2:$3}' |
awk 'BEGIN{ print "<style>table,th,td {border:1px solid black;}

Sample output from kubectl:

NAMESPACE   NAME      READY       STATUS              RESTARTS
ABC       ABC-jkij     1/1        RUNNING          897
BAC       BAC-jkij     2/2        RUNNING          897
HJI       HJI-jkij     2/2        RUNNING          897

kubectl get pods | grep -v 1/1 | grep -v 2/2 | awk '{print $1:$2:$3}'

The above command will result only the headers like below as i kept -v:

NAMESPACE   NAME      READY       STATUS              RESTARTS

so in Awk we have no of records variable right, I want to put a condition in the Awk block which should print only if there are any 0/1 or 0/2 pod results, not just the headers.

So to summmarize

NAMESPACE   NAME      READY       STATUS              RESTARTS

--> awk should not print anything

NAMESPACE   NAME      READY       STATUS              RESTARTS
ABC       ABC-jkij     0/1        RUNNING               897
ABC       ABC-jkij     0/2        RUNNING               897

Awk should print only in above scenarios.

Above command will give the headers like NAME NAMESPACE STATUS as output if there are no pods in with 0/1 0/2 etc., status. Now I want to include an if condition saying if (NR >1 ) then only it should print else it should not print anything ie., headers. If I am trying to put if (NR>1) in begin block it's still printing the headers.

Upvotes: 0

Views: 467

Answers (1)

tripleee
tripleee

Reputation: 189377

You should generally avoid piping Awk into Awk; it's a scripting language, so you can put as many statements as you want.

kubectl get pods | 
awk 'NR>1 && !/1\/1|2\/2/ {
    if (!headers++) print "<style>table,th,td {border:1px solid black;}</style><table>"
    print "<tr>"
    for (i=1; i<=3; ++i) printf "<td>%s</td>\n", $i
    print "</tr>"
  }
  END { if (headers) print "</table>" }'

I had to guess a bit what your expected output would be but if there are glitches, I expect it should be easy to see what to fix.

Upvotes: 3

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